POJ-3017 Cut the Sequence DP+单调队列+堆

　　这题的DP方程是容易想到的，f[i]=Min{ f[j]+Max(num[j+1],num[j+2],......,num[i]) | 满足m的下界<j<=i }，复杂度O(n^2)，妥妥的TLE。其实很多都决策都是没有必要的，只要保存在满足m的区间内，num值单调递减的的那些决策。如果遍历的话，一个下降的序列会退化到O(n^2)，于是用堆来优化。。。堆优化这里，纠结了很久T_T，，，网上很多代码都是直接用set来处理，但是set在erase元素的都是会把相同的元素都除掉，应该是只erase一个元素，因为相同的元素中其它的可能会存在队列中。。。难道是数据弱了？。。。

 1 //STATUS:C++_AC_1172MS_1352KB
 2 #include <functional>
 3 #include <algorithm>
 4 #include <iostream>
 5 //#include <ext/rope>
 6 #include <fstream>
 7 #include <sstream>
 8 #include <iomanip>
 9 #include <numeric>
10 #include <cstring>
11 #include <cassert>
12 #include <cstdio>
13 #include <string>
14 #include <vector>
15 #include <bitset>
16 #include <queue>
17 #include <stack>
18 #include <cmath>
19 #include <ctime>
20 #include <list>
21 #include <set>
22 #include <map>
23 using namespace std;
24 //using namespace __gnu_cxx;
25 //define
26 #define pii pair<int,int>
27 #define mem(a,b) memset(a,b,sizeof(a))
28 #define lson l,mid,rt<<1
29 #define rson mid+1,r,rt<<1|1
30 #define PI acos(-1.0)
31 //typedef
32 typedef __int64 LL;
33 typedef unsigned __int64 ULL;
34 //const
35 const int N=100010;
36 const int INF=0x3f3f3f3f;
37 const int MOD=100000,STA=8000010;
38 const LL LNF=1LL<<60;
39 const double EPS=1e-8;
40 const double OO=1e15;
41 const int dx[4]={-1,0,1,0};
42 const int dy[4]={0,1,0,-1};
43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
44 //Daily Use ...
45 inline int sign(double x){return (x>EPS)-(x<-EPS);}
46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
49 template<class T> inline T Min(T a,T b){return a<b?a:b;}
50 template<class T> inline T Max(T a,T b){return a>b?a:b;}
51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
55 //End
56 
57 int num[N],q[N];
58 int n;
59 LL m,f[N];
60 multiset<int> sbt;
61 
62 int main()
63 {
64  //   freopen("in.txt","r",stdin);
65     int i,j,l,r,p,ok;
66     LL sum;
67     while(~scanf("%d%I64d",&n,&m))
68     {
69         l=sum=0;r=-1;
70         sbt.clear();
71         ok=1;
72         for(i=p=1;i<=n;i++){
73             scanf("%d",&num[i]);
74             sum+=num[i];
75             while(sum>m)sum-=num[p++];
76             if(p>i){ok=0;break;}
77             while(l<=r && num[i]>=num[q[r]]){
78                 if(l<r)sbt.erase(f[q[r-1]]+num[q[r]]);
79                 r--;
80             }
81             q[++r]=i;
82             if(l<r)sbt.insert(f[q[r-1]]+num[q[r]]);
83             while(q[l]<p){
84                 if(l<r)sbt.erase(f[q[l]]+num[q[l+1]]);
85                 l++;
86             }
87             f[i]=f[p-1]+num[q[l]];
88             if(l<r)f[i]=Min(f[i],(LL)*sbt.begin());
89         }
90         for(;i<=n;i++)
91             scanf("%d",&j);
92 
93         printf("%I64d
",ok?f[n]:-1);
94     }
95     return 0;
96 }